We consider boundary value problem for nonlinear fractional differential equation 𝐷𝛼0+𝑢(𝑡)+𝑓(𝑡,𝑢(𝑡))=0,0<𝑡<1,𝑛−1<𝛼≤𝑛,𝑛>3,𝑢(0)=𝑢′(1)=ğ‘¢î…žî…ž(0)=⋯=𝑢(𝑛−1)(0)=0, where 𝐷𝛼0+ denotes the Caputo fractional derivative. By using fixed point theorem, we obtain some new results for the existence and multiplicity of solutions to a higher-order fractional boundary value problem. The interesting point lies in the fact that the solutions here are positive, monotone, and concave.

1. Introduction

In this paper, we deal with the following boundary value problem for higher-order fractional differential equation: 𝐷𝛼0+𝑢(𝑡)+𝑓(𝑡,𝑢(𝑡))=0,0<𝑡<1,𝑛−1<𝛼⩽𝑛,𝑛>3,𝑢(0)=ğ‘¢î…ž(1)=ğ‘¢î…žî…ž(0)=⋯=𝑢(𝑛−1)(0)=0,(1.1) where 𝐷𝛼0+ denotes the Caputo fractional derivative and 𝑓∶[0,1]×[0,+∞)→[0,+∞) is a real function. By using fixed point theorem, some sufficient conditions for existence and multiplicity of solutions to the above boundary value problem are obtained. Moreover, we will show that the solutions obtained here are positive, monotone, and concave.

Fractional differential equations are valuable tools in the modelling of many phenomena in various fields of science and engineering [1–5]. Due to their applications, fractional differential equations have gained considerable attentions and there has been a significant development in the study of existence of solutions, and positive solutions to boundary value problems for fractional differential equations (e.g., [6–9] and references therein).

Some papers are devoted to study the existence of solutions for higher-order fractional boundary value problem. Salem [10] investigated the existence of pseudosolutions for the nonlinear 𝑚-point boundary value problem of fractional type 𝐷𝛼]𝑥(𝑡)+ğ‘ž(𝑡)𝑓(𝑡,𝑥(𝑡))=0,0<𝑡<1,𝛼∈(𝑛−1,𝑛,𝑛⩾2,𝑥(0)=ğ‘¥î…ž(0)=ğ‘¥î…žî…ž(0)=⋯=𝑥(𝑛−2)(0)=0,𝑥(1)=𝑚−2𝑖=1𝜉𝑖𝑥𝜂𝑖.(1.2) Zhang [11] considered the existence of positive solutions to the singular boundary value problem for fractional differential equations 𝐷𝛼0+𝑢(𝑡)+ğ‘ž(𝑡)𝑓𝑢,ğ‘¢î…ž,…,𝑢(𝑛−2)]=0,0<𝑡<1,𝛼∈(𝑛−1,𝑛,𝑛⩾2,𝑢(0)=ğ‘¢î…žî…ž(0)=⋯=𝑢(𝑛−2)(0)=𝑢(𝑛−2)(1)=0,(1.3) where 𝐷𝛼0+ is the Riemann-Liouville fractional derivative of order 𝛼. In another paper, Zhang [9] studied the existence, multiplicity, and nonexistence of positive solutions for the following higher-order fractional boundary value problem: 𝐷𝛼]𝑢𝑢+ğœ†â„Ž(𝑡)𝑓(𝑢)=0,0<𝑡<1,𝛼∈(𝑛−1,𝑛,𝑛⩾2,(1)=ğ‘¢î…ž(0)=⋯=𝑢(𝑛−2)(0)=𝑢(𝑛−1)(0)=0,(1.4) where 𝐷𝛼 is the Caputo fractional derivative of order 𝛼.

It seems that the authors of the papers only studied the existence of the solutions or positive solutions. No one consider the qualities of the solutions for boundary value problems of fractional differential equation. Motivated by all the above works, the aim of this paper is to study the monotone, concave, and positive solutions of a fractional differential equation.

The rest of the paper is organized as follows. In Section 2, we will introduce some lemmas and definitions which will be used later. In Section 3, the existence and multiplicity of positive solutions for the boundary value problem (1.1) will be discussed. In Section 4, examples are given to check our results.

2. Basic Definitions and Preliminaries

In this section, we introduce some necessary definitions and lemmas, which will be used in the proofs of our main results.

Definition 2.1 (see [12]). The integral 𝐼𝛼0+1𝑦(𝑡)=Γ(𝛼)𝑡0(𝑡−𝑠)𝛼−1𝑦(𝑠)𝑑𝑠,(2.1) where 𝛼>0 is called the Riemann-Liouville fractional integral of order 𝛼.

Definition 2.2 (see [12]). The Caputo fractional derivative for a function 𝑦∶(0,∞)→𝑅 can be written as 𝐷𝛼0+1𝑦(𝑡)=Γ(𝑛−𝛼)𝑡0(𝑡−𝑠)𝑛−𝛼−1𝑦(𝑛)(𝑠)𝑑𝑠,(2.2) where 𝑛=[𝛼]+1, [𝛼] denotes the integer part of real number 𝛼.

According to the definitions of fractional calculus, we can obtain that the fractional integral and the Caputo fractional derivative satisfy the following Lemma.

Lemma 2.3 (see [13]). Assume that 𝑢∈𝐶𝑚[0,1] and 𝜌∈(𝑚−1,𝑚),𝑚∈𝑁 and 𝑣∈𝐶1[0,1]. Then, for 𝑡∈[0,1], (a)𝐷𝜌0+𝐼𝜌0+𝑣(𝑡)=𝑣(𝑡), (b)𝐼𝜌0+𝐷𝜌0+∑𝑢(𝑡)=𝑢(𝑡)−𝑚−1𝑘=0((𝑢(𝑘)(0))/𝑘!)𝑡𝑘, (c)lim𝑡→0+𝐷𝜌0+𝑢(𝑡)=lim𝑡→0+𝐼𝜌0+𝑢(𝑡)=0.

Definition 2.4. Let 𝐸 be a real Banach space over 𝑅. A nonempty convex closed set 𝑝⊂𝐸 is said to be a cone, provided that(a)ğ‘Žğ‘¢âˆˆğ‘ƒ,forall𝑢∈𝑃,ğ‘Žâ©¾0, (b)𝑢,−𝑢∈𝑃,implies𝑢=0.

Definition 2.5. Let 𝐸 be a real Banach space and 𝑃⊂𝐸 a cone. A function 𝜑∶𝑃→[0,∞) is called a nonnegative continuous concave functional if 𝜑 is continuous and 𝜑(𝜆𝑥+(1−𝜆)𝑦)⩾𝜆𝜑(𝑥)+(1−𝜆)𝜑(𝑦),(2.3) for all 𝑥,𝑦∈𝑃 and 0⩽𝜆⩽1.

Lemma 2.6 (see [14]). Let 𝐸 be a Banach space, 𝐾⊆𝐸 a cone in 𝐸, and Ω1, Ω2 two bounded open subsets of 𝐸 with 0∈Ω1 and Ω1⊂Ω2. Suppose that 𝑇∶𝐾∩(Ω2⧵Ω1)→𝐾 is continuous and completely continuous such that either (i)‖𝑇𝑢‖⩽‖𝑢‖for𝑢∈𝐾∩𝜕Ω1,‖𝑇𝑢‖⩾‖𝑢‖for𝑢∈𝐾∩𝜕Ω2,or (ii)‖𝑇𝑢‖⩾‖𝑢‖for𝑢∈𝐾∩𝜕Ω1,‖𝑇𝑢‖⩽‖𝑢‖for𝑢∈𝐾∩𝜕Ω2holds. Then, 𝑇 has a fixed point in 𝐾∩(Ω2⧵Ω1).
Let 𝑏,𝑑,𝑟>0 be constants, 𝑃𝑟={𝑢∈𝑃∶‖𝑢‖<𝑟}, 𝑃(𝜑,𝑏,𝑑)={𝑢∈𝑃∶𝑏⩽𝜑(𝑢),‖𝑢‖⩽𝑑}.

Lemma 2.7 (see [15]). Let 𝑃 be a cone in real Banach space 𝐸. Let 𝑇∶𝑃𝑐→𝑃𝑐 be a completely continuous map and 𝜑 a nonnegative continuous concave functional on 𝑃 such that 𝜑(𝑢)⩽‖𝑢‖, for all 𝑢∈𝑃𝑐. Suppose that there exist constants ğ‘Ž,𝑏,𝑑 with 0<ğ‘Ž<𝑏<𝑑⩽𝑐 such that ‖(i){𝑢∈𝑃(𝜑,𝑏,𝑑)∶𝜑(𝑢)>𝑏}≠∅,𝜑(𝑇𝑢)>𝑏∀𝑢∈𝑃(𝜑,𝑏,𝑑),(ii)𝑇𝑢‖<ğ‘Žâˆ€ğ‘¢âˆˆğ‘ƒğ‘Ž,(iii)𝜑(𝑇𝑢)>𝑏,∀𝑢∈𝑃(𝜑,𝑏,𝑐)with‖𝑇𝑢‖>𝑑.(2.4) Then, 𝑇 has at least three fixed points 𝑢1, 𝑢2, and 𝑢3 satisfying ‖‖𝑢1‖‖𝑢<ğ‘Ž,𝑏<𝜑2,‖‖𝑢3‖‖𝑢>ğ‘Ž,𝜑3<𝑏.(2.5)

Lemma 2.8. Assume that 𝑓(𝑡,𝑢)∈𝐶([0,1]×[0,+∞),[0,+∞)), then 𝑢∈𝐶[0,1] be a solution of fractional boundary value problem (1.1) if and only if 𝑢∈𝐶[0,1] is a solution of integral equation 𝑢(𝑡)=10𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠,(2.6) where ⎧⎪⎨⎪⎩𝐺(𝑡,𝑠)=(𝛼−1)𝑡(1−𝑠)𝛼−2−(𝑡−𝑠)𝛼−1Γ(𝛼),0⩽𝑠⩽𝑡⩽1,(𝛼−1)𝑡(1−𝑠)𝛼−2Γ(𝛼),0⩽𝑡⩽𝑠⩽1.(2.7)

Proof. Firstly, we prove the necessity. Let 𝑢∈𝐶[0,1] is a solution of fractional boundary value problem (1.1). By Lemma 2.3, we have 𝑢(𝑡)=−𝐼𝛼0+𝑓(𝑡,𝑢(𝑡))+𝑢(0)+ğ‘¢î…žğ‘¢(0)𝑡+⋯+(𝑛−1)(0)(𝑡𝑛−1)!𝑛−11=−Γ(𝛼)𝑡0(𝑡−𝑠)𝛼−1𝑓(𝑠,𝑢(𝑠))𝑑𝑠+ğ‘¢î…ž(0)𝑡.(2.8) Therefore, ğ‘¢î…ž(1𝑡)=−Γ(𝛼)𝑡0(𝛼−1)(𝑡−𝑠)𝛼−2𝑓(𝑠,𝑢(𝑠))𝑑𝑠+ğ‘¢î…ž(0).(2.9) By the boundary value condition ğ‘¢î…ž(1)=0, we have ğ‘¢î…ž1(0)=Γ(𝛼)10(𝛼−1)(1−𝑠)𝛼−2𝑓(𝑠,𝑢(𝑠))𝑑𝑠.(2.10) Hence, we obtain 1𝑢(𝑡)=−Γ(𝛼)𝑡0(𝑡−𝑠)𝛼−11𝑓(𝑠,𝑢(𝑠))𝑑𝑠+Γ(𝛼)10(𝛼−1)𝑡(1−𝑠)𝛼−2=𝑓(𝑠,𝑢(𝑠))𝑑𝑠10𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠.(2.11) The necessity is proved.
Now, we prove the sufficiency. Let 𝑢∈𝐶[0,1] be a solution of integral equation (2.6). Then, we have 𝐷𝛼0+𝑢(𝑡)=−𝐷𝛼0+𝑡0(𝑡−𝑠)𝛼−1+Γ(𝛼)𝑓(𝑠,𝑢(𝑠))𝑑𝑠10(𝛼−1)(1−𝑠)𝛼−2𝐷Γ(𝛼)𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝛼0+𝑡=−𝐷𝛼0+𝐼𝛼0+𝑓(𝑡,𝑢(𝑡))=−𝑓(𝑡,𝑢(𝑡)).(2.12) By direct computation, we obtain that 𝑢(0)=ğ‘¢î…ž(1)=ğ‘¢î…žî…ž(0)=⋯=𝑢(𝑛−1)(0)=0.(2.13) That is to say, 𝑢 is a solution of fractional boundary value problem (1.1). Thus, the sufficiency is proved.

Lemma 2.9. Let 𝑓(𝑡,𝑢(𝑡))∈𝐶([0,1]×[0,∞),[0,∞)). Then, the solution 𝑢(𝑡) of fractional boundary value problem (1.1) satisfies (1)𝑢(𝑡)isconcaveon(0,1), (2)𝑢(𝑡)⩾0isincreasingfor𝑡∈[0,1].

Proof. Suppose that 𝑢(𝑡) is a solution of fractional boundary value problem (1.1). By (2.11), we know that 1𝑢(𝑡)=−Γ(𝛼)𝑡0(𝑡−𝑠)𝛼−11𝑓(𝑠,𝑢(𝑠))𝑑𝑠+Γ(𝛼)10(𝛼−1)𝑡(1−𝑠)𝛼−2𝑓(𝑠,𝑢(𝑠))𝑑𝑠.(2.14) Therefore, ğ‘¢î…ž1(𝑡)=−Γ(𝛼)𝑡0(𝛼−1)(𝑡−𝑠)𝛼−21𝑓(𝑠,𝑢(𝑠))𝑑𝑠+Γ(𝛼)10(𝛼−1)(1−𝑠)𝛼−2𝑢𝑓(𝑠,𝑢(𝑠))𝑑𝑠,(2.15)(1𝑡)=−Γ(𝛼)𝑡0(𝛼−1)(𝛼−2)(𝑡−𝑠)𝛼−3[]𝑓(𝑠,𝑢(𝑠))𝑑𝑠⩽0,for𝑡∈0,1,𝑛−1<𝛼⩽𝑛,𝑛>3,(2.16) which implies that 𝑢(𝑡) is concave on (0,1). The statement (1) is proved.
Since ğ‘¢î…žî…ž(𝑡)⩽0, we know that 𝑢′(𝑡) is nonincreasing. By ğ‘¢î…ž(1)=0, we have ğ‘¢î…ž(𝑡)⩾0, 𝑡∈[0,1]. Thus, 𝑢(𝑡) is increasing. Noting 𝑢(0)=0, we obtain that 𝑢(𝑡)⩾0 for 𝑡∈[0,1]. The statement (2) is proved.

Lemma 2.10. The Green's function 𝐺(𝑡,𝑠), defined by (2.7), satisfies (1)max0⩽𝑡⩽1𝐺(𝑡,𝑠)=𝐺(1,𝑠),𝑠∈[0,1], (2)𝐺(𝑡,𝑠)⩾0,𝑡,𝑠∈[0,1], (3)min𝜉⩽𝑡⩽𝜂𝐺(𝑡,𝑠)⩾𝜉𝛼−1𝐺(1,𝑠),𝑠∈[0,1],forall𝜉,𝜂∈(0,1),𝜉<𝜂.

Proof. By (2.7), we have ğºî…žğ‘¡âŽ§âŽªâŽ¨âŽªâŽ©(𝑡,𝑠)=(𝛼−1)(1−𝑠)𝛼−2−(𝛼−1)(𝑡−𝑠)𝛼−2Γ(𝛼),0⩽𝑠⩽𝑡⩽1,(𝛼−1)(1−𝑠)𝛼−2Γ(𝛼),0⩽𝑡⩽𝑠⩽1.(2.17) It is clear that ğºî…žğ‘¡(𝑡,𝑠)⩾0,𝑡,𝑠∈[0,1]. Therefore, 𝐺(𝑡,𝑠) is increasing respect to 𝑡 for 𝑠∈[0,1]. Thus, max0⩽𝑡⩽1𝐺(𝑡,𝑠)=𝐺(1,𝑠). The statement (1) holds.
If 0⩽𝑡⩽𝑠⩽1, then, 𝐺(𝑡,𝑠)=(𝛼−1)𝑡(1−𝑠)𝛼−2Γ(𝛼)⩾0.(2.18) Since 𝐺(𝑡,𝑠) is increasing respect to 𝑡 for 𝑠∈[0,1], it is easy to see that 𝐺(𝑡,𝑠)⩾0 for 0⩽𝑠⩽𝑡⩽1. We get the statement (2).
On the other hand, we have minğœ‰â©½ğ‘¡â©½ğœ‚âŽ§âŽªâŽªâŽªâŽ¨âŽªâŽªâŽªâŽ©ğº(𝑡,𝑠)=(𝛼−1)𝜉(1−𝑠)𝛼−2−(𝜉−𝑠)𝛼−1[],Γ(𝛼),𝑠∈0,𝜉(𝛼−1)𝜉(1−𝑠)𝛼−2[],(Γ(𝛼),𝑠∈𝜉,𝜂𝛼−1)𝜉(1−𝑠)𝛼−2[]=⎧⎪⎨⎪⎩(Γ(𝛼),𝑠∈𝜂,1𝛼−1)𝜉(1−𝑠)𝛼−2−(𝜉−𝑠)𝛼−1[],Γ(𝛼),𝑠∈0,𝜉(𝛼−1)𝜉(1−𝑠)𝛼−2[].Γ(𝛼),𝑠∈𝜉,1(2.19) If 𝑠∈[0,𝜉], then (𝛼−1)𝜉(1−𝑠)𝛼−2−(𝜉−𝑠)𝛼−1=𝜉(𝛼−1)(1−𝑠)𝛼−2−𝜉𝛼−1(1−(𝑠/𝜉))𝛼−1⩾𝜉𝛼−1(𝛼−1)(1−𝑠)𝛼−2−𝜉𝛼−1(1−𝑠)𝛼−1=𝜉𝛼−1(𝛼−1)(1−𝑠)𝛼−2−(1−𝑠)𝛼−1.(2.20) If 𝑠∈[𝜉,1], then (𝛼−1)𝜉(1−𝑠)𝛼−2⩾(𝛼−1)𝜉𝛼−1(1−𝑠)𝛼−2⩾𝜉𝛼−1(𝛼−1)(1−𝑠)𝛼−2−(1−𝑠)𝛼−1.(2.21) Thus, min𝜉⩽𝑡⩽𝜂𝐺(𝑡,𝑠)⩾𝜉𝛼−1(𝛼−1)(1−𝑠)𝛼−2−(1−𝑠)𝛼−1Γ(𝛼)=𝜉𝛼−1[].𝐺(1,𝑠),𝑠∈0,1(2.22) This yields the statement (3). The proof is finished.

3. Main Results

In this section, we establish the results for the existence and multiplicity of monotone and concave positive solutions for fractional boundary value problem (1.1).

Let 𝐸=𝐶[0,1] with ‖𝑢‖=max0⩽𝑡⩽1|𝑢(𝑡)|. We define the cone 𝑃⊂𝐸 by []𝑃=𝑢∈𝐸∶𝑢(𝑡)⩾0isconcaveon0,1,min𝜉⩽𝑡⩽𝜂𝑢(𝑡)⩾𝜉𝛼−1‖𝑢‖.(3.1)

And denote the operator 𝑇 by 𝑇𝑢(𝑡)=10𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠.(3.2)

Lemma 3.1. Assume that 𝑓(𝑡,𝑢)∈𝐶([0,1]×[0,+∞),[0,+∞)), then 𝑇∶𝑃→𝑃 is completely continuous.

Proof. In view of non-negativeness and continuity of 𝐺(𝑡,𝑠) and 𝑓(𝑡,𝑢(𝑡)), we know that the operator 𝑇 is continuous and 𝑇𝑢(𝑡)⩾0, for 𝑢∈𝑃.
By (2.16), we have (𝑇𝑢)(1𝑡)=−Γ(𝛼)𝑡0(𝛼−1)(𝛼−2)(𝑡−𝑠)𝛼−3[]𝑓(𝑠,𝑢(𝑠))𝑑𝑠⩽0,for𝑡∈0,1,𝑛−1<𝛼⩽𝑛,𝑛>3.(3.3)
Moreover, it follows from Lemma 2.10 that for 𝑢∈𝑃, min𝜉⩽𝑡⩽𝜂𝑇𝑢(𝑡)=min𝜉⩽𝑡⩽𝜂10𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠⩾𝜉𝛼−110𝐺(1,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠=𝜉𝛼−1‖𝑇𝑢‖.(3.4) Therefore, the operator 𝑇∶𝑃→𝑃 is well defined.
Assume that Ω∈𝑃 is bounded; that is, there exists a positive constant 𝑀>0 such that ‖𝑢‖⩽𝑀 for all 𝑢∈Ω. Let 𝑁=max0⩽𝑡⩽1,‖𝑢‖⩽𝑀|𝑓(𝑡,𝑢(𝑡))|+1. For all 𝑢∈Ω, we have ||||=||||𝑇𝑢(𝑡)10||||𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠⩽𝑁10𝐺(1,𝑠)𝑑𝑠,(3.5) which shows that 𝑇(Ω) is uniformly bounded.
In addition, for each 𝑢∈Ω, 𝑡1,𝑡2∈[0,1] such that 𝑡1<𝑡2, we have ||𝑡𝑇𝑢2𝑡−𝑇𝑢1||=||||10𝐺𝑡2,𝑠𝑓(𝑠,𝑢(𝑠))𝑑𝑠−10𝐺𝑡1||||⩽|||||,𝑠𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑡10𝑡1−𝑠𝛼−1−𝑡2−𝑠𝛼−1Γ(𝛼)𝑓(𝑠,𝑢(𝑠))𝑑𝑠−𝑡2𝑡1𝑡2−𝑠𝛼−1|||||+||||Γ(𝛼)𝑓(𝑠,𝑢(𝑠))𝑑𝑠10𝑡(𝛼−1)2−𝑡1(1−𝑠)𝛼−2𝑓||||⩽𝑁Γ(𝛼)(𝑠,𝑢(𝑠))𝑑𝑠𝑡Γ(𝛼+1)𝛼2−𝑡𝛼1+𝑁𝑡Γ(𝛼)2−𝑡1.(3.6) Thus, by the standard arguments, we obtain that 𝑇(Ω) is equicontinuous. The Arzela-Ascoli theorem implies that 𝑇∶𝑃→𝑃 is completely continuous. The proof is completed.

Let 𝑀1=10𝐺(1,𝑠)𝑑𝑠,𝑁1=𝜉𝛼−1𝜂𝜉𝐺(1,𝑠)𝑑𝑠.(3.7)

Theorem 3.2. Let 𝑓(𝑡,𝑢)∈𝐶([0,1]×[0,∞),[0,∞)). Assume that there exist two positive constants 𝑟2>𝑟1>0 such that (𝐻1)𝑓(𝑡,𝑢)≤𝑟2/𝑀1for(𝑡,𝑢)∈[0,1]×[0,𝑟2], (𝐻2)𝑓(𝑡,𝑢)≥𝑟1/𝑁1for(𝑡,𝑢)∈[0,1]×[0,𝑟1]. Then, fractional boundary value problem (1.1) has at least one positive, increasing, and concave solution 𝑢 such that 𝑟1≤‖𝑢‖≤𝑟2.

Proof. Lemmas 2.8 and 3.1 imply that 𝑇∶𝑃→𝑃 is completely continuous and fractional boundary problem (1.1) has a solution 𝑢=𝑢(𝑡) if and only if 𝑢 satisfies the operator equation 𝑢=𝑇𝑢.
Let Ω1∶={𝑢∈𝑃∶‖𝑢‖<𝑟1}. By (𝐻2), for 𝑢∈𝜕Ω1 and 𝑡∈[0,1], we have 𝑇𝑢(𝑡)=10⩾𝑟𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠1𝑁1𝜂𝜉min𝜉≤𝑡≤𝜂⩾𝑟𝐺(𝑡,𝑠)𝑑𝑠1𝑁1𝜉𝛼−1𝜂𝜉𝐺(1,𝑠)𝑑𝑠=𝑟1=‖𝑢‖.(3.8) So, ‖𝑇𝑢‖⩾‖𝑢‖,for𝑢∈𝜕Ω1.(3.9)
Let Ω2∶={𝑢∈𝑃∶‖𝑢‖<𝑟2}. For 𝑢∈𝜕Ω2, and 𝑡∈[0,1], it follows from (𝐻1) that ‖𝑇𝑢(𝑡)‖=max0≤𝑡≤1||||10||||⩽𝑟𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠2𝑀110=𝑟𝐺(1,𝑠)𝑑𝑠2𝑀110𝐺(1,𝑠)𝑑𝑠=𝑟2=‖𝑢‖.(3.10) Lemma 2.6 implies that the fractional boundary value problem (1.1) has at least one positive solution 𝑢 such that 𝑟1≤‖𝑢‖≤𝑟2. By Lemma 2.9, the solution is also increasing and concave.

In order to use Lemma 2.7, we define the nonnegative continuous concave functional 𝜑 by 𝜑(𝑢)=min𝜉⩽𝑡⩽𝜂𝑢(𝑡),forall𝑢∈𝑃.

Theorem 3.3. Suppose that 𝑓(𝑡,𝑢)∈𝐶([0,1]×[0,∞),[0,∞)) and there exist constants 0<ğ‘Ž<𝑏<𝑐 such that the following conditions hold: (𝐻3)ğ‘Žğ‘“(𝑡,𝑢)<𝑀1for(𝑡,𝑢)∈[0,1]×[0,ğ‘Ž],(𝐻4)𝑏𝑓(𝑡,𝑢)>𝑁1for(𝑡,𝑢)∈[𝜉,𝜂]×[𝑏,𝑐],(𝐻5)𝑐𝑓(𝑡,𝑢)⩽𝑀1for(𝑡,𝑢)∈[0,1]×[0,𝑐].Then, the fractional boundary problem (1.1) has at least three positive, increasing, and concave solutions 𝑢1, 𝑢2, and 𝑢3 such that max0⩽𝑡⩽1||𝑢1||(𝑡)<ğ‘Ž,𝑏<min𝜉⩽𝑡⩽𝜂||𝑢2||(𝑡)<max0⩽𝑡⩽1||𝑢2||(𝑡)⩽𝑐,ğ‘Ž<max0⩽𝑡⩽1||𝑢3||(𝑡)⩽𝑐,min𝜉⩽𝑡⩽𝜂||𝑢3||(𝑡)<𝑏.(3.11)

Proof. By Lemmas 2.8 and 3.1, 𝑇∶𝑃→𝑃 is completely continuous, and fractional boundary value problem (1.1) has a solution 𝑢=𝑢(𝑡) if and only if 𝑢 satisfies the operator equation 𝑢=𝑇𝑢.
First of all, we will prove the following assertions.
Assertion 3.4 (𝑇(𝑃𝑐)⊆𝑃𝑐 and 𝑇(ğ‘ƒğ‘Ž)âŠ†ğ‘ƒğ‘Ž). Firstly, Lemma 3.1 guarantees 𝑇(𝑃𝑐)⊆𝑃. Secondly, for all 𝑢∈𝑃𝑐, we have ‖𝑢‖⩽𝑐. By (𝐻5), ‖𝑇𝑢(𝑡)‖=max0≤𝑡≤1||||10||||⩽𝑐𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑀110𝐺(1,𝑠)𝑑𝑠=𝑐,(3.12) which implies that 𝑇(𝑃𝑐)⊆𝑃𝑐. In the same way, 𝑇(ğ‘ƒğ‘Ž)âŠ†ğ‘ƒğ‘Ž.Assertion 3.5 ({𝑢∈𝑃(𝜑,𝑏,𝑑)∣𝜑(𝑢)>𝑏}≠∅, and 𝜑(𝑇𝑢)>𝑏, for all 𝑢∈𝑃(𝜑,𝑏,𝑑)). Let 𝑑=𝑐, 𝑢=(𝑏+𝑐)/2. Then, ‖𝑢‖<𝑑 and 𝜑(𝑢)=𝜑((𝑏+𝑐)/2)>𝑏. Consequently, {𝑢∈𝑃(𝜑,𝑏,𝑑)∣𝜑(𝑢)>𝑏}≠∅.
If 𝑢∈𝑃(𝜑,𝑏,𝑑), then from (𝐻4) and Lemma 2.10, we obtain that 𝜑(𝑇𝑢)=min𝜉⩽𝑡⩽𝜂⩾𝑇𝑢(𝑡)𝜂𝜉min𝜉⩽𝑡⩽𝜂𝐺>𝑏(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑁1𝜉𝛼−1𝜂𝜉𝐺(1,𝑠)𝑑𝑠=𝑏.(3.13) That is, 𝜑(𝑇𝑢)>𝑏, for all 𝑢∈𝑃(𝜑,𝑏,𝑑).
Assertion 3.6 (𝜑(𝑢)>𝑏, for all 𝑢∈𝑃(𝜑,𝑏,𝑐) with ‖𝑢‖>𝑑). If 𝑢∈𝑃(𝜑,𝑏,𝑐) and ‖𝑇𝑢‖>𝑑=𝑐, similar to the above, we also have 𝜑(𝑇𝑢)>𝑏.
Assertions 1∼3 imply that all conditions of Lemma 2.7 hold. Therefore, the fractional boundary value problem (1.1) has at least three positive solutions 𝑢1,𝑢2, and 𝑢3 satisfying max0⩽𝑡⩽1||𝑢1||(𝑡)<ğ‘Ž,𝑏<min𝜉⩽𝑡⩽𝜂||𝑢2||(𝑡)<max0⩽𝑡⩽1||𝑢2||(𝑡)⩽𝑐,ğ‘Ž<max0⩽𝑡⩽1||𝑢3||(𝑡)⩽𝑐,min𝜉⩽𝑡⩽𝜂||𝑢3||(𝑡)<𝑏.(3.14) By Lemma 2.9, the positive solutions are also increasing and concave. The proof is completed.

4. Example

In this section, we will present some examples to show the effectiveness of our work.

Example 4.1. Consider the fractional boundary value problem 𝐷7/20+𝑢(𝑡)+𝑡2+√𝑢+4=0,0<𝑡<1,𝑢(0)=ğ‘¢î…žî…ž(0)=ğ‘¢î…žî…žî…ž(0)=ğ‘¢î…ž(1)=0.(4.1) Setting 𝜉=1/2,𝜂=3/4, we obtain 𝑀1=10𝐺(1,𝑠)𝑑𝑠=(7/2)−1Γ(7/2)10(1−𝑠)(7/2)−21𝑑𝑠−Γ(7/2)10(1−𝑠)(7/2)−1=1𝑑𝑠−1(5/2)Γ(5/2)𝑁(7/2)Γ(7/2)≈0.2149,1=12(7/2)−13/41/2=𝐺(1,𝑠)𝑑𝑠(1/2)5/2Γ(7/2)3/41/252(1−𝑠)(7/2)−2𝑑𝑠−3/41/2(1−𝑠)(7/2)−1𝑑𝑠≈0.0065.(4.2) Let 𝑟1=1/40,𝑟2=4. We have 𝑓(𝑡,𝑢)=𝑡2+√𝑟𝑢+4⩽2𝑀1[]×[],≈18.613,for(𝑡,𝑢)∈0,10,4𝑓(𝑡,𝑢)=𝑡2+√𝑟𝑢+4⩾1𝑁1[]×1≈3.846,for(𝑡,𝑢)∈0,10,.40(4.3) Theorem 3.2 implies that fractional boundary value problem (4.1) has at least one positive, increasing, and concave solution. The approximate solution is obtained by the Adams-type predictor-corrector method [16], which is displayed in Figure 1 for the step size ℎ=0.01.

Example 4.2. Consider the fractional boundary value problem 𝐷7/20+𝑢(𝑡)+𝑓(𝑡,𝑢)=0,0<𝑡<1,𝑢(0)=ğ‘¢î…žî…ž(0)=ğ‘¢î…žî…žî…ž(0)=ğ‘¢î…ž(1)=0,(4.4) where âŽ§âŽªâŽ¨âŽªâŽ©ğ‘¡ğ‘“(𝑡,𝑢)=10+155𝑢3[],𝑡,0⩽𝑢⩽1,𝑡∈0,1[].10+𝑢+154,𝑢>1,𝑡∈0,1(4.5)
Setting 𝜉=1/2,𝜂=3/4, we know that 𝑀1≈0.2149,𝑁1≈0.0065. Choosing ğ‘Ž=1/10,𝑏=1,𝑐=45, we obtain that 𝑡𝑓(𝑡,𝑢)=10+155𝑢3<ğ‘Žğ‘€1[]×1≈0.466for(𝑡,𝑢)∈0,10,,𝑡10𝑓(𝑡,𝑢)=𝑏10+𝑢+154>𝑁11≈153.846for(𝑡,𝑢)∈2,34×[],𝑡1,45𝑓(𝑡,𝑢)=𝑐10+𝑢+154<𝑀1[]×[].≈209.4for(𝑡,𝑢)∈0,10,45(4.6) Theorem 3.3 implies that fractional boundary value problem (4.4) has three positive, increasing, and concave solutions such that max0⩽𝑡⩽1||𝑢1||<1(𝑡)10,1<min1/2⩽𝑡⩽3/4||𝑢2||(𝑡)<max0⩽𝑡⩽1||𝑢2||1(𝑡)⩽45,10<max0⩽𝑡⩽1||𝑢3||(𝑡)⩽45,min1/2⩽𝑡⩽3/4||𝑢3||(𝑡)<1.(4.7) For numerical simulation case 1, Figure 2 depicts the phase responses state variables of 𝑢(𝑡) with the step size ℎ=0.01.


This work was jointly supported by Natural Science Foundation of China (no. 10871214), Natural Science Foundation of Hunan Provincial under Grants nos. 11JJ3005, 10JJ6007, and 2010GK3008.